The maximum value of frac{log x}{x} in (2, in | vedclass

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(C) Let $y = \frac{\log x}{x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac

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$\frac{1}{e}$

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(C) Let $y = \frac{\log x}{x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.For maxima,set $\frac{dy}{dx} = 0$:$\frac{1 - \log x}{x^2} = 0 \implies 1 - \log x = 0 \implies \log x = 1 \implies x = e$.Since $e \approx 2.718$,$x = e$ lies in the interval $(2, \infty)$.Now,check the second derivative:$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2x \log x - 3x}{x^4}$.At $x = e$,$\frac{d^2y}{dx^2} = \frac{2e(1) - 3e}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} Since the second derivative is negative,the function has a maximum at $x = e$.The maximum value is $y = \frac{\log e}{e} = \frac{1}{e}$.

The maximum value of $\frac{\log x}{x}$ in $(2, \infty)$ is

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